22 Inequalities and MVT for Integrals, Fundamental Theorem of Calculus, Integration Techniques (by Parts, Substitution)

Lecture from: 15.05.2024 | Video: Video ETHZ

Review: Who Gets to be Integrated?

Quick Recap: Integrable Functions

• Continuous functions on closed, bounded intervals $[a,b]$ are always Riemann integrable. (This is a big one!)
• Monotonic functions (always increasing or always decreasing) on closed, bounded intervals are also always Riemann integrable, even if they have some jumps.
• The integral is linear: ${\int }_a^b({\lambda }_1{f}_1(x)+{\lambda }_2{f}_2(x))dx={\lambda }_1{\int }_a^b{f}_1(x)dx+{\lambda }_2{\int }_a^b{f}_2(x)dx$. This means we can break down integrals of sums and pull out constant factors, just like with derivatives.

Today’s Agenda:

• Inequalities for integrals.
• The Mean Value Theorem for Integrals.
• The big one: The Fundamental Theorem of Calculus (also known as “Hauptsatz der Differential- und Integralrechnung”), which shows the amazing connection between differentiation and integration.

Clicker Question: An Integral Involving an Odd Function

What is the value of ${\int }_{-1}^1\sin (x{)}^{11}dx$?

Answer: 0

Reasoning

The function $f(x)=\sin (x{)}^{11}$ is an odd function. This means $f(-x)=(\sin (-x){)}^{11}=(-\sin (x){)}^{11}=-(\sin (x){)}^{11}=-f(x)$.

When you integrate an odd function over a symmetric interval like $[-a,a]$ (here, $[-1,1]$), the integral is always zero.

Why? Because the “negative area” (where the function is below the x-axis) on one side perfectly cancels out the “positive area” (where the function is above the x-axis) on the other side.

We can write: ${\int }_{-1}^1\sin (x{)}^{11}dx={\int }_{-1}^0\sin (x{)}^{11}dx+{\int }_0^1\sin (x{)}^{11}dx$ Let $u=-x$ in the first integral. Then $du=-dx$. When $x=-1,u=1$. When $x=0,u=0$.

${\int }_{-1}^0\sin (x{)}^{11}dx={\int }_1^0\sin (-u{)}^{11}(-du)={\int }_1^0(-\sin (u){)}^{11}(-du)$ $={\int }_1^0-(\sin (u){)}^{11}(-du)={\int }_1^0(\sin (u){)}^{11}du=-{\int }_0^1(\sin (u){)}^{11}du$. (We flipped the limits of integration, which introduces a minus sign).

So, ${\int }_{-1}^1\sin (x{)}^{11}dx=-{\int }_0^1\sin (x{)}^{11}dx+{\int }_0^1\sin (x{)}^{11}dx=0$ (We don’t even need to know the value of ${\int }_0^1\sin (x{)}^{11}dx$ for this problem!)

Inequalities and the Mean Value Theorem for Integrals

Theorem: Monotonicity of the Integral

If One Function is Bigger, Its Integral is Bigger This is a very intuitive property. If you have two integrable functions, $f(x)$ and $g(x)$, and $f(x)$ is always less than or equal to $g(x)$ on an interval $[a,b]$, then the area under $f(x)$ must be less than or equal to the area under $g(x)$ on that same interval.

Let $f,g:[a,b]\to \mathbb{R}$ be bounded and Riemann integrable. If $f(x)\le g(x)$ for all $x\in [a,b]$, then: ${\int }_a^{b}f(x)dx\le {\int }_a^{b}g(x)dx$ This is often called the “monotonicity of the integral”.

Proof Idea

Consider the difference $h(x)=g(x)-f(x)$. Since $f(x)\le g(x)$, we have $h(x)\ge 0$ for all $x\in [a,b]$.

The integral ${\int }_a^{b}h(x)dx$ represents the area above $f$ and below $g$, which must be non-negative.

Using linearity: ${\int }_a^b(g(x)-f(x))dx={\int }_a^{b}g(x)dx-{\int }_a^{b}f(x)dx$. Since ${\int }_a^{b}h(x)dx\ge 0$, we have ${\int }_a^{b}g(x)dx-{\int }_a^{b}f(x)dx\ge 0$, which gives the result.

Corollary: Triangle Inequality for Integrals

The Integral of the Absolute Value is Greater The absolute value of an integral is less than or equal to the integral of the absolute value of the function. This is like the triangle inequality $|a+b|\le |a|+|b|$ but for integrals (which are essentially infinite sums).

Let $f:[a,b]\to \mathbb{R}$ be bounded and Riemann integrable. Then: $\left|{\int }_a^{b}f(x)dx\right|\le {\int }_a^b|f(x)|dx$

Proof

We know that for any real number $y$, $-|y|\le y\le |y|$.

So, for our function $f(x)$, we have $-|f(x)|\le f(x)\le |f(x)|$ for all $x\in [a,b]$.

Since $|f|$ is also integrable if $f$ is, we can apply the monotonicity theorem twice:

1. $f(x)\le |f(x)|⟹{\int }_a^{b}f(x)dx\le {\int }_a^b|f(x)|dx$.
2. $-|f(x)|\le f(x)⟹{\int }_a^b-|f(x)|dx\le {\int }_a^{b}f(x)dx⟹-{\int }_a^b|f(x)|dx\le {\int }_a^{b}f(x)dx$.

Combining these two, we have $-{\int }_a^b|f(x)|dx\le {\int }_a^{b}f(x)dx\le {\int }_a^b|f(x)|dx$.

This is precisely the definition of $\left|{\int }_a^{b}f(x)dx\right|\le {\int }_a^b|f(x)|dx$.

Theorem: Cauchy-Schwarz Inequality for Integrals

A Quick Note on Integrals of Products: Remember from the last lecture’s clicker question, there’s no simple general inequality between $\int (f\cdot g)$ and $(\int f)\cdot (\int g)$ like there is for sums. However, there is a famous inequality involving integrals of products…

A Powerful Inequality for Products Inside Integrals This inequality gives an upper bound for the absolute value of the integral of a product $f(x)g(x)$ in terms of the integrals of $f(x{)}^2$ and $g(x{)}^2$. It’s a continuous analog of the Cauchy-Schwarz inequality for sums/vectors.

Let $f,g:[a,b]\to \mathbb{R}$ be bounded and Riemann integrable. Then: $\left|{\int }_a^{b}f(x)g(x)dx\right|\le \sqrt{{\int }_a^b(f(x){)}^{2}dx}\cdot \sqrt{{\int }_a^b(g(x){)}^{2}dx}$ (This can be written as $|B|\le \sqrt{A\cdot C}$ using the notation in the proof below).

Proof

From our theorem on properties of integrable functions, if $f$ and $g$ are integrable, then so are $f^2$, $g^2$, and $fg$.

Consider the function $(f(x)-t\cdot g(x){)}^2$ for any real number $t$. Since it’s a square, $(f(x)-t\cdot g(x){)}^2\ge 0$ for all $x$.

By monotonicity of the integral, ${\int }_a^b(f(x)-t\cdot g(x){)}^{2}dx\ge 0$. Let’s expand the integrand: $(f(x){)}^2-2tf(x)g(x)+t^2(g(x){)}^2$.

Using linearity of the integral: $\underset{A}{\underbrace{{\int }_a^b(f(x){)}^{2}dx}}-2t\underset{B}{\underbrace{{\int }_a^{b}f(x)g(x)dx}}+t^2\underset{C}{\underbrace{{\int }_a^b(g(x){)}^{2}dx}}\ge 0$ So we have a quadratic in $t$: $A-2Bt+C{t}^2\ge 0$ for all $t\in \mathbb{R}$.

Let $P(t)=C{t}^2-2Bt+A\ge 0$.

• Case 1: $C=0$. This means ${\int }_a^b(g(x){)}^{2}dx=0$. If $g$ were continuous, this would imply $g(x)=0$ everywhere, but for general integrable functions, it means $g(x)$ is “zero almost everywhere” in a measure theory sense. For our purposes, if $g(x{)}^2\ge 0$ and its integral is 0, this forces $g(x)$ to be essentially zero in terms of its contribution to integrals.

  If $C=0$, the inequality becomes $A-2Bt\ge 0$ for all $t\in \mathbb{R}$. This is a linear function of $t$. For it to be always non-negative, its slope (which is $-2B$) must be zero. So, $B=0$.

  If $C=0$ and $B=0$, then the Cauchy-Schwarz inequality $|B|\le \sqrt{AC}$ becomes $0\le \sqrt{A\cdot 0}$, which is $0\le 0$. True.

• Case 2: $C>0$. Then $P(t)=C{t}^2-2Bt+A$ is an upward-opening parabola. For this parabola to be always greater than or equal to zero (i.e., never dip below the t-axis), it can have at most one real root. This means its discriminant must be less than or equal to zero. Discriminant: $(-2B{)}^2-4CA\le 0$. $4B^2-4AC\le 0⟹4B^2\le 4AC⟹B^2\le AC$. Taking the square root (since $A,C\ge 0$), we get $|B|\le \sqrt{AC}$. This is exactly $\left|{\int }_a^{b}fgdx\right|\le \sqrt{{\int }_a^b{f}^{2}dx}\sqrt{{\int }_a^b{g}^{2}dx}$.

• Case 3: $C<0$. This cannot happen because $C={\int }_a^b(g(x){)}^{2}dx$ and $(g(x){)}^2\ge 0$, so its integral must be $\ge 0$.

This establishes the inequality.

Conventions for Integral Limits

A Few Handy Rules for Integral Boundaries These conventions make working with integrals more flexible, especially when dealing with additive properties or changing variables.

(Compare with Remark 5.2.9 in the script)

1. Integral over a point: ${\int }_a^{a}f(x)dx=0$. (The “area” of a line is zero).
2. Flipping the limits: If $a<b$, then ${\int }_b^{a}f(x)dx=-{\int }_a^{b}f(x)dx$.

With these conventions, the following property holds for any $a,b,c\in \mathbb{R}$ (where $f$ is integrable on the largest interval spanned by $a,b,c$):

Additivity of the interval of integration: ${\int }_a^{c}f(x)dx={\int }_a^{b}f(x)dx+{\int }_b^{c}f(x)dx$ (Note: $a<b<c$ is not required here due to the conventions above).

Theorem: Mean Value Theorem for Integrals

The “Average Value” of a Continuous Function If $f$ is continuous on $[c,d]$, this theorem says there’s some point $\xi$ within that interval such that the value of the function at that point, $f(\xi )$, is exactly the average value of the function over the interval. The average value is defined as $\frac{1}{d-c}{\int }_c^{d}f(x)dx$. Geometrically, it means there’s a rectangle with height $f(\xi )$ and width $d-c$ that has the exact same area as the area under the curve $f(x)$ from $c$ to $d$.

Let $f:[a,b]\to \mathbb{R}$ be continuous. Then for any $c,d\in [a,b]$ with $c\le d$, there exists a $\xi \in [c,d]$ such that: ${\int }_c^{d}f(x)dx=f(\xi )(d-c)$

Proof

Without loss of generality, assume $c<d$. (If $c=d$, both sides are 0, and any $\xi \in [c,c]$ works).

Since $f$ is continuous on the closed, bounded interval $[c,d]$, by the Min-Max Theorem (Extreme Value Theorem), $f$ attains its minimum value $m=f(u)$ and maximum value $M=f(v)$ on $[c,d]$ for some $u,v\in [c,d]$.

So, $m\le f(x)\le M$ for all $x\in [c,d]$.

By the monotonicity of the integral: ${\int }_c^{d}mdx\le {\int }_c^{d}f(x)dx\le {\int }_c^{d}Mdx$. $m(d-c)\le {\int }_c^{d}f(x)dx\le M(d-c)$.

Dividing by $(d-c)$ (which is positive): $m\le \frac{1}{d-c}{\int }_c^{d}f(x)dx\le M$ Let $Y=\frac{1}{d-c}{\int }_c^{d}f(x)dx$. We have shown $Y$ is a value between the minimum ($m=f(u)$) and maximum ($M=f(v)$) values of the continuous function $f$ on $[c,d]$.

By the Intermediate Value Theorem (for continuous functions), since $Y$ is between $f(u)$ and $f(v)$, there must exist some $\xi \in [c,d]$ such that $f(\xi )=Y$.

So, $f(\xi )=\frac{1}{d-c}{\int }_c^{d}f(x)dx$, which means ${\int }_c^{d}f(x)dx=f(\xi )(d-c)$.

Example (showing continuity is important)

Consider $f:[0,1]\to \mathbb{R}$, $f(x)=\{\begin{array}{ll}0& 0\le x<1/2\\ 1& 1/2\le x\le 1\end{array}$.

This function is not continuous at $x=1/2$.

${\int }_0^{1}f(x)dx={\int }_0^{1/2}0dx+{\int }_{1/2}^{1}1dx=0+(1-1/2)=1/2$.

So we are looking for $\xi \in [0,1]$ such that $f(\xi )(1-0)=1/2$, i.e., $f(\xi )=1/2$. However, $f(x)$ only takes values $0$ or $1$. There is no $\xi \in [0,1]$ such that $f(\xi )=1/2$.

This shows that continuity is an important assumption in the Mean Value Theorem for Integrals.

The Fundamental Theorem of Calculus (FTC)

The Bridge Between Derivatives and Integrals! This is arguably one of the most important theorems in all of calculus. It establishes the incredible inverse relationship between differentiation and integration.

Part 1 essentially says: If you define a function $F(x)$ as the “area so far” under $f(t)$ (from some fixed point $c$ up to $x$), then the rate of change of this area function $F(x)$ is precisely the original function $f(x)$. In other words, $\frac{d}{dx}\left({\int }_c^{x}f(t)dt\right)=f(x)$.

Part 2 then gives us a powerful way to calculate definite integrals: if you can find any antiderivative $F$ of $f$ (meaning $F^{\prime }=f$), then ${\int }_a^{b}f(x)dx=F(b)-F(a)$.

“The integral is the inverse operation of differentiation.”

Theorem: First Fundamental Theorem of Calculus (FTC Part 1)

Let $c\in [a,b]$ and $f:[a,b]\to \mathbb{R}$ be continuous. Define the “area function” (or “integral function”) $F:[a,b]\to \mathbb{R}$ by: $F(x)={\int }_c^{x}f(t)dt(\text{for }x\in [a,b])$ Then $F$ is continuously differentiable on $[a,b]$ and its derivative is $f$: $F^{\prime }(x)=f(x)\text{for all }x\in [a,b]$

Proof

Let $x_0,x\in [a,b]$. Using the additive property of integrals: ${\int }_c^{x}f(t)dt={\int }_c^{x_0}f(t)dt+{\int }_{x_0}^{x}f(t)dt$.

So, $F(x)=F(x_0)+{\int }_{x_0}^{x}f(t)dt$. This means $F(x)-F(x_0)={\int }_{x_0}^{x}f(t)dt$.

By the Mean Value Theorem for Integrals (since $f$ is continuous), there exists a ${\xi }_x$ between $x_0$ and $x$ such that ${\int }_{x_0}^{x}f(t)dt=f({\xi }_x)(x-x_0)$.

So, $F(x)-F(x_0)=f({\xi }_x)(x-x_0)$. For $x\ne x_0$, divide by $(x-x_0)$: $\frac{F(x)-F(x_0)}{x-x_0}=f({\xi }_x)$ Now, we take the limit as $x\to x_0$. As $x\to x_0$, ${\xi }_x$ (which is squeezed between $x_0$ and $x$) must also approach $x_0$.

Since $f$ is continuous at $x_0$, ${\lim }_{x\to x_0}f({\xi }_x)={\lim }_{{\xi }_x\to x_0}f({\xi }_x)=f(x_0)$.

Therefore, $F^{\prime }(x_0)={\lim }_{x\to x_0}\frac{F(x)-F(x_0)}{x-x_0}={\lim }_{x\to x_0}f({\xi }_x)=f(x_0)$ Since $x_0$ was arbitrary in $[a,b]$, $F^{\prime }(x)=f(x)$ for all $x\in [a,b]$. Since $f$ is continuous, $F^{\prime }$ is continuous, so $F$ is continuously differentiable. $\square$

Definition: Antiderivative (Stammfunktion)

Let $f:[a,b]\to \mathbb{R}$ be continuous. A function $F:[a,b]\to \mathbb{R}$ is called an antiderivative (or “Stammfunktion”) of $f$ if $F$ is (continuously) differentiable on $[a,b]$ and $F^{\prime }(x)=f(x)$ for all $x\in [a,b]$.

Theorem: Second Fundamental Theorem of Calculus (FTC Part 2) & Uniqueness of Antiderivatives

Calculating Definite Integrals Easily! FTC Part 1 told us that ${\int }_c^{x}f(t)dt$ is an antiderivative of $f(x)$. This part says:

1. Antiderivatives for a given $f$ are almost unique – they only differ by a constant.
2. If you can find any antiderivative $F$ of $f$, then the definite integral ${\int }_c^{d}f(x)dx$ is simply $F(d)-F(c)$. This is incredibly powerful for computations!

Let $f:[a,b]\to \mathbb{R}$ be continuous.

1. There exists an antiderivative of $f$. (FTC Part 1 showed that $G(x)={\int }_a^{x}f(t)dt$ is one such antiderivative).
2. Any antiderivative of $f$ is uniquely determined up to an additive constant. (If $F_1$ and $F_2$ are both antiderivatives of $f$, then $F_1(x)=F_2(x)+C$ for some constant $C$).
3. For any antiderivative $F$ of $f$, and for any $c,d\in [a,b]$: ${\int }_c^{d}f(x)dx=F(d)-F(c)$ This is often written as ${F(x)|}_c^d=F(d)-F(c)$. (”$F$ evaluated from $c$ to $d$”).

Proof

• Existence (Part 1): We proved in FTC Part 1 (Theorem 5.4.1) that $G(x)={\int }_a^{x}f(t)dt$ is an antiderivative of $f$.

• Uniqueness up to a constant (Part 2): Let $F_1$ and $F_2$ be two antiderivatives of $f$. Then $F_1^{\prime }(x)=f(x)$ and $F_2^{\prime }(x)=f(x)$. Consider the function $H(x)=F_1(x)-F_2(x)$. Then $H^{\prime }(x)=F_1^{\prime }(x)-F_2^{\prime }(x)=f(x)-f(x)=0$ for all $x\in [a,b]$. Since $H^{\prime }(x)=0$, by a corollary of the MVT (Corollary 4.2.5 part 1), $H(x)$ must be a constant, say $C$. So, $F_1(x)-F_2(x)=C$, which means $F_1(x)=F_2(x)+C$.

• Evaluation Formula (Part 3): Let $F$ be any antiderivative of $f$. We know from Part 2 that $F(x)$ must be of the form $G(x)+C$, where $G(x)={\int }_a^{x}f(t)dt$ is the specific antiderivative from FTC Part 1 (with starting point $a$). So, $F(x)={\int }_a^{x}f(t)dt+C$ for some constant $C$.

  Now, let’s evaluate $F(d)-F(c)$: $F(d)-F(c)=\left({\int }_a^{d}f(t)dt+C\right)-\left({\int }_a^{c}f(t)dt+C\right)$ $={\int }_a^{d}f(t)dt+C-{\int }_a^{c}f(t)dt-C$ $={\int }_a^{d}f(t)dt-{\int }_a^{c}f(t)dt$ Using the property ${\int }_a^d={\int }_a^c+{\int }_c^d$, we have ${\int }_a^d-{\int }_a^c={\int }_c^d$. So, $F(d)-F(c)={\int }_c^{d}f(t)dt$.

Clicker Question: Derivative of an Integral with Variable Upper Limit

Let $f:\mathbb{R}\to \mathbb{R}$ be continuous and define $G(x)={\int }_0^{x^2}f(t)dt$. What is $G^{\prime }(x)$?

Let $H(u)={\int }_0^{u}f(t)dt$. By FTC Part 1, $H^{\prime }(u)=f(u)$. Our function is $G(x)=H(x^2)$. This is a composition.

We need to use the Chain Rule: $G^{\prime }(x)=H^{\prime }(x^2)\cdot \frac{d}{dx}(x^2)$. $H^{\prime }(x^2)=f(x^2)$ (since $H^{\prime }(u)=f(u)$). $\frac{d}{dx}(x^2)=2x$. So, $G^{\prime }(x)=f(x^2)\cdot 2x=2xf(x^2)$.

Common Mistake: To just say $G^{\prime }(x)=f(x^2)$. This forgets to apply the chain rule for the upper limit $x^2$. The “inside function” $u(x)=x^2$ needs to be differentiated.

The Fundamental Theorem of Calculus reduces the problem of calculating definite integrals to the problem of finding an antiderivative.

Examples: Using Antiderivatives to Calculate Integrals

1. $f(x)=x$. An antiderivative is $F(x)=x^2/2$. ${\int }_0^{a}xdx={\frac{x^2}{2}|}_0^a=\frac{a^2}{2}-\frac{0^2}{2}=\frac{a^2}{2}$ This is the area of a triangle with base $a$ and height $a$. More generally, ${\int }_a^{b}xdx={\frac{x^2}{2}|}_a^b=\frac{b^2}{2}-\frac{a^2}{2}=\frac{b^2-a^2}{2}$ (This matches our direct calculation from Example 5.1.5).

2. $f(x)=x^2$. An antiderivative is $F(x)=x^3/3$. ${\int }_0^a{x}^{2}dx={\frac{x^3}{3}|}_0^a=\frac{a^3}{3}-\frac{0^3}{3}=\frac{a^3}{3}$ More generally, ${\int }_a^b{x}^{2}dx={\frac{x^3}{3}|}_a^b=\frac{b^3}{3}-\frac{a^3}{3}=\frac{b^3-a^3}{3}$

Techniques of Integration

The FTC is powerful, but it relies on us being able to find an antiderivative. Two common techniques help with this:

Integration by Parts

The Product Rule for Derivatives, in Reverse! If you remember the product rule for derivatives: $(u(x)v(x){)}^{\prime }=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$. If we integrate both sides, we get $u(x)v(x)=\int u^{\prime }(x)v(x)dx+\int u(x)v^{\prime }(x)dx$.

Rearranging this gives the formula for integration by parts. It’s useful when you have a product of functions to integrate, and differentiating one part and integrating the other makes the new integral simpler.

Let $f,g:[a,b]\to \mathbb{R}$ be continuously differentiable. Then: ${\int }_a^{b}f(x)g^{\prime }(x)dx={f(x)g(x)|}_a^b-{\int }_a^b{f}^{\prime }(x)g(x)dx$ $=(f(b)g(b)-f(a)g(a))-{\int }_a^b{f}^{\prime }(x)g(x)dx$

Proof

Let $H(x)=f(x)g(x)$. $H(x)$ is continuously differentiable, and by the product rule: $H^{\prime }(x)=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$.

So, $H(x)$ is an antiderivative of $f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$.

By FTC Part 2: ${\int }_a^b(f^{\prime }(x)g(x)+f(x)g^{\prime }(x))dx=H(b)-H(a)=f(b)g(b)-f(a)g(a)$.

Using linearity of the integral: ${\int }_a^b{f}^{\prime }(x)g(x)dx+{\int }_a^{b}f(x)g^{\prime }(x)dx=f(b)g(b)-f(a)g(a)$.

Rearranging gives the desired formula.

Integration by Substitution (Change of Variables)

The Chain Rule for Derivatives, in Reverse! Substitution is like undoing the chain rule. If part of your integrand looks like the result of a chain rule differentiation (an “inside function” $\varphi (t)$ and its derivative ${\varphi }^{\prime }(t)$ appearing), substitution can simplify the integral.

Let $\varphi :[a,b]\to \mathbb{R}$ be continuously differentiable. Let $I\subseteq \mathbb{R}$ be an interval such that the image $\varphi ([a,b])\subseteq I$. Let $g:I\to \mathbb{R}$ be continuous. Then: ${\int }_a^{b}g(\varphi (t)){\varphi }^{\prime }(t)dt={\int }_{\varphi (a)}^{\varphi (b)}g(x)dx$

Proof Idea

Let $F$ be an antiderivative of $g$ on $I$. (It exists because $g$ is continuous on $I$, by FTC Part 1). So $F^{\prime }(x)=g(x)$.

Consider the composite function $(F\circ \varphi )(t)=F(\varphi (t))$. Its derivative, by the Chain Rule, is $(F\circ \varphi {)}^{\prime }(t)=F^{\prime }(\varphi (t)){\varphi }^{\prime }(t)=g(\varphi (t)){\varphi }^{\prime }(t)$.

So, $F(\varphi (t))$ is an antiderivative of $g(\varphi (t)){\varphi }^{\prime }(t)$.

By FTC Part 2, applied to the left-hand side integral: ${\int }_a^{b}g(\varphi (t)){\varphi }^{\prime }(t)dt={F(\varphi (t))|}_a^b=F(\varphi (b))-F(\varphi (a))$.

Now consider the right-hand side integral. Since $F$ is an antiderivative of $g$: ${\int }_{\varphi (a)}^{\varphi (b)}g(x)dx={F(x)|}_{\varphi (a)}^{\varphi (b)}=F(\varphi (b))-F(\varphi (a))$.

Both sides are equal.

Continue here: 23 Area of Circle, Volume of Sphere, Indefinite Integrals, Partial Fraction Decomposition